国开电大 复变函数 形考任务3答案.pdf

国开电大复变函数形考任务3答案一、选择题1.02.03.cos1+isin1eiei4.2i二、填空题e1e5.26.e(cos1+isin1)i7.21Ln2i(2k)8.24k为整数三、计算题9.解:z2(xiy)2x2y22xyi∴ez2ex2y2i2xyexp[(x2y2)[cos(2xy)isin(2xy)]]ez2∴=ex2y2exp(z2)=ex2y210.解:∵z=x+iy11xyi∴zxiyx2y2x2y2xyy1(cosisin)∴ezex2y2x2y2x2y21xyRe(e2)ex2y2cos∴x2y211.解:∵lnz=i/2∴由对数函数的定义有:ei/2cosisiniZ=22∴所给方程的解为z=iez13i2(cosisin)12.解:∵33eLn2(cosisin)=33根据指数函数的定义有:z=Ln(1+3i)四、证明题13.证明：根据正弦函数及余弦正数定义有：e2ize2izsin2z2ieizizeizeiz2sinzcosz22i2e2ize2iz2i∴sin2z=2sinz·cosz14.证明:令A=1cosxcos2xcosnxB=sinx+sin2x+…sinnx∴ABi1eixei2xeinxn11eizx1ei(n1)x2ixx1ei21e2n1n1ixn122isinxesinxn22ixe2xxxi2isine2sin22n1sinxnn2(cosxisinx)xsin22=2n1sinxnsinxsin2xsinx2sinxxsin2∴2